Factor x^4 - x^2 + 1/4.

Step-by-step solution

1. 1

   Treat $x^2$ as the working variable. Only even powers appear, and $x^4=\left(x^2\right)^2$, so the expression is a quadratic in $u=x^2$:

   $u^2-u+\frac14$

2. 2

   Check the perfect-square pattern with the fractional constant. $\tfrac14=\left(\tfrac12\right)^2$, and the middle term must equal $-2ab$:

   $-2\cdot u\cdot\frac12=-u$

   which matches. A fraction in the constant is no obstacle — the identity works over any field.

3. 3

   Write the square and substitute back.

   $u^2-u+\frac14=\left(u-\frac12\right)^2\quad\Longrightarrow\quad x^4-x^2+\frac14=\left(x^2-\frac12\right)^2$

4. 4

   Decide whether to go further. Over the reals you could split $x^2-\tfrac12$ as a difference of squares:

   $\left(x-\tfrac{1}{\sqrt2}\right)^2\left(x+\tfrac{1}{\sqrt2}\right)^2$

   but over the rationals $\left(x^2-\tfrac12\right)^2$ is as far as it goes, since $\tfrac12$ is not a rational square.

5. 5

   Verify numerically. At $x=2$: the original is $16-4+0.25=12.25$, and $\left(4-0.5\right)^2=3.5^2=12.25$ $\checkmark$.